Question 7 - A-Level Maths

Edexcel C4 — Question 7 12 marks

Exam Board	Edexcel
Module	C4 (Core Mathematics 4)
Marks	12
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Differentiating Transcendental Functions
Type	Differentiate general exponentials
Difficulty	Standard +0.8 Part (a) requires proving the derivative formula for general exponentials (a standard A-level proof). Part (b) involves applying this to find a tangent equation with logarithmic coefficients. Part (c) requires solving 4^x·ln4 = 2^(x-1)·ln2 for a stationary point, which involves manipulating exponentials and logarithms to find an exact answer. The multi-step nature, proof requirement, and algebraic manipulation of transcendental functions make this moderately challenging but still within standard C4 scope.
Spec	1.07m Tangents and normals: gradient and equations 1.07n Stationary points: find maxima, minima using derivatives

7. (a) Prove that $$\frac { \mathrm { d } } { \mathrm {~d} x } \left( a ^ { x } \right) = a ^ { x } \ln a .$$ A curve has the equation $y = 4 ^ { x } - 2 ^ { x - 1 } + 1$.
(b) Show that the tangent to the curve at the point where it crosses the $y$-axis has the equation $$3 x \ln 2 - 2 y + 3 = 0 .$$ (c) Find the exact coordinates of the stationary point of the curve.
7. continued

Show mark scheme Show mark scheme source

Answer	Marks	Guidance
(a) let $y = a^x \therefore \ln y = x \ln a$	M1
$\frac{1}{y}\frac{dy}{dx} = \ln a$	M1
$\frac{dy}{dx} = y \ln a = a^x \ln a \quad \therefore \frac{d}{dx}(a^x) = a^x \ln a$	A1
(b) $\frac{dy}{dx} = 4^x \ln 4 - 2^{x-1} \ln 2$	M1 A1
$x = 0, y = \frac{3}{2}, \quad \text{grad} = \ln 4 - \frac{1}{2}\ln 2 = \frac{3}{2}\ln 2$	M1
$\therefore y - (\frac{3}{2}\ln 2)x + \frac{3}{2}, \quad 2y = 3x \ln 2 + 3, \quad 3x \ln 2 - 2y + 3 = 0$	M1 A1
(c) $4^x \ln 4 - 2^{x-1} \ln 2 = 0$	M1
$(2^x)^2 \times 2 \ln 2 - \frac{1}{2}(2^x) \ln 2 = 0$	M1
$\frac{1}{2}(2^x)\ln 2[4(2^x) - 1] = 0$	M1
$2^x = \frac{1}{4}, \quad x = -2 \quad (-2, -\frac{15}{16})$	A2	(12)

**(a)** let $y = a^x \therefore \ln y = x \ln a$ | M1 |

$\frac{1}{y}\frac{dy}{dx} = \ln a$ | M1 |

$\frac{dy}{dx} = y \ln a = a^x \ln a \quad \therefore \frac{d}{dx}(a^x) = a^x \ln a$ | A1 |

**(b)** $\frac{dy}{dx} = 4^x \ln 4 - 2^{x-1} \ln 2$ | M1 A1 |

$x = 0, y = \frac{3}{2}, \quad \text{grad} = \ln 4 - \frac{1}{2}\ln 2 = \frac{3}{2}\ln 2$ | M1 |

$\therefore y - (\frac{3}{2}\ln 2)x + \frac{3}{2}, \quad 2y = 3x \ln 2 + 3, \quad 3x \ln 2 - 2y + 3 = 0$ | M1 A1 |

**(c)** $4^x \ln 4 - 2^{x-1} \ln 2 = 0$ | M1 |

$(2^x)^2 \times 2 \ln 2 - \frac{1}{2}(2^x) \ln 2 = 0$ | M1 |

$\frac{1}{2}(2^x)\ln 2[4(2^x) - 1] = 0$ | M1 |

$2^x = \frac{1}{4}, \quad x = -2 \quad (-2, -\frac{15}{16})$ | A2 | (12)

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7. (a) Prove that

$$\frac { \mathrm { d } } { \mathrm {~d} x } \left( a ^ { x } \right) = a ^ { x } \ln a .$$

A curve has the equation $y = 4 ^ { x } - 2 ^ { x - 1 } + 1$.\\
(b) Show that the tangent to the curve at the point where it crosses the $y$-axis has the equation

$$3 x \ln 2 - 2 y + 3 = 0 .$$

(c) Find the exact coordinates of the stationary point of the curve.\\

7. continued\\

\hfill \mbox{\textit{Edexcel C4  Q7 [12]}}

This paper (8 questions)

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Q1 4 Q2 8 Q3 9 Q4 9 Q5 10 Q6 10 Q7 12 Q8 13

Answer	Marks	Guidance
(a) let \(y = a^x \therefore \ln y = x \ln a\)	M1
\(\frac{1}{y}\frac{dy}{dx} = \ln a\)	M1
\(\frac{dy}{dx} = y \ln a = a^x \ln a \quad \therefore \frac{d}{dx}(a^x) = a^x \ln a\)	A1
(b) \(\frac{dy}{dx} = 4^x \ln 4 - 2^{x-1} \ln 2\)	M1 A1
\(x = 0, y = \frac{3}{2}, \quad \text{grad} = \ln 4 - \frac{1}{2}\ln 2 = \frac{3}{2}\ln 2\)	M1
\(\therefore y - (\frac{3}{2}\ln 2)x + \frac{3}{2}, \quad 2y = 3x \ln 2 + 3, \quad 3x \ln 2 - 2y + 3 = 0\)	M1 A1
(c) \(4^x \ln 4 - 2^{x-1} \ln 2 = 0\)	M1
\((2^x)^2 \times 2 \ln 2 - \frac{1}{2}(2^x) \ln 2 = 0\)	M1
\(\frac{1}{2}(2^x)\ln 2[4(2^x) - 1] = 0\)	M1
\(2^x = \frac{1}{4}, \quad x = -2 \quad (-2, -\frac{15}{16})\)	A2	(12)